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Math Help - Riemann integral

  1. #1
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    Riemann integral

    http://www.tau.ac.il/~tsirel/Courses...sis3/lect6.pdf

    Hi'I need help with the following question:


    6g11 page 86
    Thank's in advance
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  2. #2
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    Re: Riemann integral

    Where are you having trouble? I recommend doing problem 6g10 first. Show the solution to that, and we can help you further. (Hint: the answer should be very similar to integrating x from 0 to 1 without the binary expansion).
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  3. #3
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    Re: Riemann integral

    If it helps, here is the idea behind 6g10:

    \beta_k(x) = \begin{cases}1 & \lfloor 2^k x \rfloor \equiv 1\pmod{2} \\ 0 & \text{otherwise}\end{cases}

    In other words, \beta_k(x) = 1 when x \in \bigcup_{n=0}^{2^{k-1}-1} \left[\dfrac{1+2n}{2^k}, \dfrac{n+1}{2^{k-1}}\right). Then \int_{[0,1]} \beta_k(x) = \sum_{n=0}^{2^{k-1}-1} \int_{\left[ \dfrac{1+2n}{2^k},\dfrac{n+1}{2^{k-1}}\right) } 1\cdot dx (since these are the only values where the function is 1). Example: \beta_1(x) =1 when x \in \left[\dfrac{1}{2},1\right), \beta_2(x) = 1 when x \in \left[\dfrac{1}{4},\dfrac{1}{2}\right)\cup \left[\dfrac{3}{4},1\right), etc.

    Then \int_{[0,1]} \sum_{k\ge 1} \dfrac{\beta_k(x)}{2^k} = \sum_{k\ge 1} \dfrac{1}{2^k} \int_{[0,1]} \beta_k(x)
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